On the fine behaviors of the eigenvalues of the linearized Gel'fand problem and its applications (Shapes and other properties of the solutions of PDEs)

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On the fine behaviors

of

the

eigenvalues

of

the

linearized Gel’fand

problem

and

its

applications1

金沢大学理工研究域大塚浩史(Hiroshi Ohtsuka)2

Faculty of Mathematics and Physics, Institute of Science and Engineering,

Kanazawa University Abstract

The purpose of this note is to overview our recent results

concern-ing the linearlized eigenvalue problem for the Gel’fand problem. The

main result is a second order estimate for the first $m$ eigenvalues of

the linearized Gel’fand problem associated to solutions which blow-up

at $m$ points. From this information, we determine some qualitative

properties ofthe first $m$ eigenfunctions.

This is based on a joint work with Francesca Gladiali (Univ.

Sas-sari) and Massimo Grossi (Univ. Roma “La Sapienza

1 The

Gel’fand

problem

The Gel’fand problem is the following semilinear elliptic problem with

expo-nential nonlinearity:

$-\Delta u=\lambda e^{u}$ $in$ $\Omega,$ $u=0$

$on$ $\partial\Omega$

, (1.1)

where $\Omega\subset \mathbb{R}^{2}$

is a bounded domain with smooth boundary $\partial\Omega$

and $\lambda>0$

is a real parameter. This problem appears in a wide variety of areas of

mathematics such as the conformalembedding ofaflat domain into a sphere,

self-dual gauge field theories, equilibrium states of large number of vortices,

stationary states of chemotaxis motion, and so forth. See [7, 8] for more

about our motivation and further references.

Especially the asymptotic behavior of the solutions as $\lambda\downarrow 0$ was studied

in detail. Let $G(x, y)$ be the Green _{function of} $-\triangle$ in $\Omega$ with Dirichlet

boundary _{condition. We divide the Green function} into two parts as usual:

$G(x, y)= \frac{1}{2\pi}\log|x-y|^{-1}+K(x, y)$_, _(1.2)

$K(x, y)$ is _called _{the regular part of} $G(x, y)$ and $R(x)=K(x, x)$ is the Robin

function. Using these functions we introduce a function over $\Omega^{m}$, which is

1This workis supported by Grant-in-Aid for Scientific Research (No.22540231), Japan

Society for the Promotion ofScience.

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know

as

the Hamiltonian function of $m$ vortices with equal intensities in the

theory of 2-dimensional incompressible non-viscous fluid:

$H^{m}(x_{1}, \ldots, x_{m}):=\frac{1}{2}\sum^{m}R(x_{j})+\frac{1}{2}\sum_{j\neq h}G(x_{j}, x_{h})j=11\leq j,h\leq m.$

Concerning the Gel’fand problem, the following result now

seems

to be classical:

Theorem 1.1 ([11]). Let $\{\lambda_{n}\}_{n\in IN}$ be

a

sequence

of

positive values such that $\lambda_{n}arrow 0$ as $narrow\infty$ and let _{$u_{n}=u_{n}(x)$} be a sequence

of

solutions

of

(1.1)

for

$\lambda=\lambda_{n}$. Then there exists some $m=0$, 1, 2,

$\cdots,$ $+\infty$ and, along a

sub-sequence,

$\lambda_{n}\int_{\Omega}e^{u_{n}}dxarrow 8\pi m$. (1.3)

Moreover, the following behaviors

_of

solutions appear in the limit $narrow\infty$:

(i)

_If

$m=0$, the sequence $\{u_{n}\}$ converges to $0$ uniformly in $\Omega.$

(ii)

_If

$m=+\infty$ the entire blow-up occurs, $i.e.$ $\inf_{K}u_{n}arrow+\infty$

for

any $K\Subset\Omega.$

(iii)

_If

$0<m<\infty$ _{the solutions} $\{u_{n}\}$ blow-up at $m$-points, that is, there is

$a\mathcal{S}et\mathcal{S}=\{\kappa_{1}, \cdots, \kappa_{m}\}\subset\Omega$

of

$m$ distinct points and a subsequence

of

$\{u_{n}\}$ such that $\Vert u_{n}\Vert_{L(\omega)}\infty=O(1)$

for

any $\omega\Subset\overline{\Omega}\backslash S,$

$u_{n}|_{S}arrow+\infty$ as _{$narrow\infty,$}

and

$u_{n}(x) arrow u_{\infty}(x) :=\sum_{j=1}^{m}8\pi G(x, \kappa_{j})$ (1.4)

locally uniformly in $\overline{\Omega}\backslash \{\kappa_{1}, ..., \kappa_{m}\}$. Furthermore the blow-up points

$S=\{\kappa_{1}, \cdots, \kappa_{m}\}$ satisfy

$\nabla H^{m}(\kappa_{1}, \ldots, \kappa_{m})=0$. (1.5)

We note that a blow-up sequence of solutions for given $S$ satisfying (1.5)

really exists under appropriate assumptions on $S$, see [1, 4, 5].

In this note we are concerned with more details about the case (iii) of

Theorem 1.1. In the following we always assume that $\{u_{n}\}$ is a sequence of

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2 The

linealized

eigenvalue

problem

of the

Gel’fand

problem

Our object in this note is the following eigenvalue problem:

$-\triangle v=\mu\lambda_{n}e^{u_{n}}v in\Omega, v=0 on\partial\Omega$, (2.1)

where $\{u_{n}\}$ is a _$m$-points blow-up sequence ofsolutions to (1.1). We are able

to

assume

that there exists a sequence of eigenvalues $\mu_{n}^{1}\leq\mu_{n}^{2}\leq\mu_{n}^{3}\leq\ldots.$

We denote k-th eigenfunction of (2.1) corresponding to the eigenvalue $\mu_{n}^{k}$ as

$v_{n}^{k}.$

We define a diagonal matrix $D$ $:=$ diag$[d_{1}, d_{1}, d_{2}, d_{2}, \cdots, d_{m}, d_{m}]$, where

$d_{j}$ is a constant given by

$d_{j}= \frac{1}{8}\exp\{4\pi R(\kappa_{j})+4\pi\sum_{1\leq i\leq m,i\neq j}G(\kappa_{j}, \kappa_{i})\}(>0)$. (2.2)

Previously we get the following behavior of$\mu_{n}^{k}$:

Theorem 2.1 ([8]). For$\lambda_{n}arrow 0$, it holds that

$\mu_{n}^{k}=-\frac{1}{2}\frac{1}{\log\lambda_{n}}+o(\frac{1}{\log\lambda_{n}})(arrow 0)$,

for

_{$1\leq k\leq m$}, (2.3) $\mu_{n}^{k}=1-48\pi\eta^{(2m+1-s)}\lambda_{n}+o(\lambda_{n})(arrow 1)$_,

for

$m+1\leq k$ $m+\mathcal{S}$) _{$\leq 3m,$}

$\mu_{n}^{k}>1,$ $fork\geq 3m+1$

where $\eta^{k}(k=1, \cdots, 2m)$ is the k-th eigenvalue

of

_the _matrix _{$D(HessH^{m})D$}

at $(\kappa_{1}, \cdots, \kappa_{m})$.

We use these to calculate the Morse index of $u_{n}$ for $n\gg 1$. Actually

we are able to get the following estimate easily from the above behaviors of

$\{\mu_{n}^{k}\}$:

$m+ind_{M}\{-H^{m}(\kappa_{1}, \cdots, \kappa_{m})\}\leq ind_{M}(u_{n})$, (2.4)

$ind_{M}^{*}(u_{n})\leq m+ind_{M}^{*}\{-H^{m}(\kappa_{1}, \cdots, \kappa_{m}$ (2.5)

where

$ind_{M}(u_{n})=\#\{k\in \mathbb{N};\mu_{n}^{k}<1\},$ $ind_{M}^{*}(u_{n})=\#\{k\in \mathbb{N};\mu_{n}^{k}\leq 1\}.$

are the Morse index and the augmented Morse index of $u_{n}$, respectively.

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are

the Morse index and the augmented Morse index of the $-H^{m}$, that is,

the numbers of the negative and non-positive eigenvalues ofHessian of $-H^{m}$

at $(\kappa_{1}, \cdots, \kappa_{m})$, respectively. These results are ageneralization ofthe results

in [6] obtained for the

case

$m=1.$

Recently we have refined the case $1\leq k\leq m$

as

follows:

We note that it seems diﬃcult to realize the matrix $(h_{ij})$ (and $D$) in the

case

$\# S=1$ considered in [6].

From the conclusion (ii), we are able to show that $v_{n}^{k}arrow 0$ outside the

blow-up set. On the other hand, we know that $c_{j}^{k}=0$ implies $v_{n}^{k}arrow 0$

locally uniformly near $\kappa_{j}$, see [8, Proposition 2.11]. Therefore we introduce

the following definition:

Definition 2.3. We say that an eigenfunction $v_{n}^{k}$ concentrates at _{$\kappa_{j}\in\Omega$}

if

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there exists $\kappa_{j,n}arrow\kappa_{j}$ such that

$|v_{n}^{k}(\kappa_{j,n})|\geq C>0$

for

$n$ large. (2.8)

As we

see

later, $c_{j}^{k}$ is obtained

as

the limit of $v_{n}^{k}(x_{j,n})$ as _{$narrow\infty$} for the

sequence satisfying $x_{j,n}arrow\kappa_{j}$ and $u_{n}(x_{j,n})arrow\infty$. Therefore it holds that

$v_{n}^{k}$ concentrates at

$\kappa_{j}$ if and only if $c_{j}^{k}\neq 0$, (2.9)

that is, we are able to count the number of peaks of $v_{n}^{k}$ from the number of

non-zero

components of $\mathfrak{c}^{k}$

as an application of this work.

Remark 2.4. We note that the behavior (2.7) for

some

$\mathfrak{c}^{k}\in[-1, 1]^{m}\subset$ $\mathbb{R}^{m}(c^{k}\neq 0)$ is obtained in the previous work [8, Proposition 2.5 and 2.13].

In this work we clarify the origin of $\mathfrak{c}^{k}$

from the

_fine

behavior of eigenvalues.

3 On

the scaling argument and the

behavior

of

eigenfunctions

In this section

we

sketch theproofof (2.7) and introduce the scaling argument

necessary to get it.

Fix $0<R\ll 1$ satisfying

$B_{2R}(\kappa_{i})\Subset\Omega$ for $i=1$, . . . ,_$m$ and $B_{R}(\kappa_{i})\cap B_{R}(\kappa_{j})=\emptyset$ if $i\neq j.$

Choose a sequence $\{x_{j,n}\}$ for each $\kappa_{j}\in S$ satisfying

$x_{j,n} arrow\kappa_{j}, u_{n}(x_{j,n})=\max_{RB(x_{j,n})}u_{n}(x)arrow\infty.$

From the Green representation formula and the behavior (1.4) of$u_{n}$, we get

$\frac{v_{n}^{k}(x)}{\mu_{n}^{k}}=\int_{\Omega}G(x, y)\lambda_{n}e^{u_{n}}v_{n}^{k}dy$

$= \sum_{j=1}^{m}\int_{B_{R}(x_{j,n})}G(x, y)\lambda_{n}e^{u_{n}}v_{n}^{k}dy+O(\lambda_{n})$.

Since $G(x, x_{j,n})$ is smooth far from $x_{j,n}$, Taylor’s theorem

$G(x, y)=G(x, x_{j,n})+(y-x_{j,n})\cdot\nabla_{y}G(x, x_{j,n})+s(x, \eta, y-x_{j,n})$,

$s(x, \eta, y-x_{j,n})=\frac{1}{2}\sum_{1\leq\alpha,\beta\leq 2}G_{y_{\alpha}y_{\beta}}(x, \eta)(y\neg x_{j,n})_{\alpha}(y-x_{j,n})_{\beta},$

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guarantees that

$\int_{B_{R}(x_{j,n})}G(x, y)\lambda_{n}e^{u_{n}}v_{n}^{k}dy$

$=G(x, x_{j,n}) \int_{B_{R}(x_{j,n})}\lambda_{n}e^{u_{n}}v_{n}^{k}dy+\nabla_{y}G(x, x_{j,n})\cdot\int_{B_{R}(x_{j,n})}(y-x_{j,n})\lambda_{n}e^{u_{n}}v_{n}^{k}dy$

$+ \frac{1}{2}\sum_{1\leq\alpha,\beta\leq 2}\int_{B_{R}(x_{j,n})}(y-x_{j,n})_{\alpha}(y-x_{j,n})_{\beta}G_{y_{\alpha}y_{\beta}}(x, \eta)\lambda_{n}e^{u_{n}}v_{n}^{k}dy$

$=:\gamma_{j,n}^{0}G(x, x_{j,n})+\gamma_{j,n}^{1}\cdot\nabla_{y}G(x, x_{j,n})+\gamma_{j,n}^{2}.$

So we need to

see

the behaviors of$\gamma_{j,n}^{0},$ $\gamma_{j,n}^{1}$, and $\gamma_{j,n}^{2}$ to get (2.7).

To this purpose we rescale the solution $u_{n}$ and eigenfunction $v_{n}^{k}$ around

$x_{j,n}$. Let $\delta_{j,n}$ be a parameter determined by the relation

$\lambda_{n}e^{u_{n}(x_{j,n})}\delta_{j,n}^{2}=1$ (3.1) and set $\tilde{u}_{j,n}(\tilde{x}):=u_{n}(\delta_{j,n}\tilde{x}+x_{j,n})-u_{n}(x_{j,n})$ in $B_{\frac{R}{\delta_{j,n}}}(0)$, $\tilde{v}_{j,n}^{k}(\tilde{x}):=v_{n}^{k}(\delta_{j,n}\tilde{x}+x_{j,n})$ in $B_{\tau_{j,\overline{n}}^{R}}(0)$.

Then it holds that

$-\triangle\tilde{u}_{j,n}=e^{\tilde{u}_{j,n}}$ in

$B_{\frac{R}{\delta_{j,n}}}(0)$, $\tilde{u}_{j,n}\leq\tilde{u}_{j,n}(0)=0$ in $B_{\frac{R}{\delta_{j,n}}}(O)$ (3.2)

and

$-\Delta\tilde{v}_{j,n}^{k}=\mu_{n}^{k}e^{\overline{u}_{j,n}}\tilde{v}_{j,n}^{k}$, in

$\tilde{x}\in B_{T_{j}^{\frac{R}{)n}}}(0)$,

$\Vert\tilde{v}_{j,n}^{k}\Vert_{L^{\infty}(B_{\frac{R}{\iota_{j,n}}}(0))}\leq 1$. (3.3)

We note that there exists $d_{j}>0$ such that

$\delta_{j,n}=d_{j}\lambda^{\frac{1}{n2}}+o(\lambda^{\frac{1}{n^{2}}})(arrow 0)$

(3.4)

from Y.Y.Li’s estimate ([10], see also [9, Corollary 4.3]). We also note that

$d_{j}$ is know to given by (2.2), see [7, Proposition 3.4].

Now assuming

$\mu_{n}^{k}arrow\mu_{\infty}^{k}\in \mathbb{R},$

we reach the following problem at the limit $narrow\infty$:

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and

$-\triangle V=\mu_{\infty}^{k}e^{U}V, \Vert V\Vert_{L^{\infty(\mathbb{R}^{2})}}\leq 1$. (3.6)

From Chen-Li’s result [3], we see

$U( \tilde{x})=\log\frac{1}{(1+\frac{|\tilde{x}|^{2}}{8})^{2}}$

and we get

$\tilde{u}_{j,n}(\tilde{x})arrow U(\tilde{x})$ in $C_{loc}^{2,\alpha}(\mathbb{R}^{2})$.

On the other hand, we can show

$\tilde{v}_{j,n}^{k}arrow V_{j}^{k}$ in $C_{loc}^{2,\alpha}(\mathbb{R}^{2})$

holds for a subsequence, where $V_{j}^{k}$ is a solution of $(3.6)$$([8,$ Proposition $2.2])$.

Since we know that there exists $j\in\{1, \cdots m\}$ such that $V_{j}^{k}\not\equiv 0([8$,

Propo-sition 2.11]), we

see

that $\mu_{\infty}^{k}$ is an eigenvalue of the linealized eigenvalue

problem (3.6) for (3.5). These eigenvalues (and the eigenfunctions) are

stud-ied in [6] and we know

$\mu_{\infty}^{k}=\frac{l(l+1)}{2}$ for

some

$l=0$, 1, 2, $\cdot$

In this note, we are interested in the case $\mu_{\infty}^{k}=0$ and for this case the

eigenfunction is known to

$V_{j}^{k}\equiv$ const. $c_{j}^{k})\in[-1, 1].$

Consequently we are able to confirm that

$\gamma_{j,n}^{0}=\int_{B_{R}(x_{j,n})}\lambda_{n}e^{u_{n}}v_{n}^{k}=\int_{B_{\frac{R}{\delta_{j,n}}}(0)}e^{\tilde{u}_{j,n}}\tilde{v}_{j,n}^{k}arrow c_{j}^{k}\int_{\mathbb{R}^{2}}e^{U}=8\pi c_{j}^{k}.$

Similarly we see

$\gamma_{j,n}^{1}=\int_{B_{R}(x_{j,n})}(y-x_{j,n})\lambda_{n}e^{u_{n}}v_{n}^{k}=\delta_{j,n}\int_{B_{\frac{R}{\delta_{j,n}}}(0)}\tilde{y}e^{\overline{u}_{j,n}}\tilde{v}_{j,n}^{k}$

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and, taking $\epsilon\in(0,1)$,

$| \int_{B_{R}(x_{j,n})}G_{y_{\alpha}y_{\beta}}(x, \eta)(y-x_{j,n})_{\alpha}(y-x_{j,n})_{\beta}\lambda_{n}e^{u_{n}}v_{n}^{k}dy|$

$\leq CR^{\epsilon}\int_{B_{R}(x_{j,n})}|y-x_{j,n}|^{2-\epsilon}\lambda_{n}e^{u_{n}}dy$

$=CR^{\epsilon} \delta_{j,n}^{2-\epsilon}\int_{B_{\frac{R}{0_{j,n}}}(0)}|\tilde{y}|^{2-\epsilon}e^{\tilde{u}_{j,n}}d\tilde{y}=O(\delta_{j,n}^{2-\epsilon})=o(\lambda^{\frac{1}{n^{2}}})$ .

For simplify the presentation, we omitted here

some

additional argument

necessary to the process $narrow\infty$, see [8, Proposition 2.5 and 2.13] for

details.

4 On the improvement of the behavior of the

eigenvalues

The sharp formula (2.6) is from the determination of the following constant

$L$:

$\frac{1}{\mu_{n}^{k}}=-2\log\lambda_{n}+L+o(1)$. (4.1)

Indeed, this leads

$\mu_{n}^{k}=\frac{1}{-2\log\lambda_{n}+L+o(1)}=-\frac{l}{2\log\lambda_{n}}\cdot\frac{1}{1-\frac{L+o(1)}{2\log\lambda_{n}}}$

$=- \frac{l}{2\log\lambda_{n}}\{1+\frac{L+o(1)}{2\log\lambda_{n}}+o(\frac{1}{\log\lambda_{n}})\}$

$=- \frac{l}{2\log\lambda_{n}}-\frac{L}{4}\cdot\frac{1}{(\log\lambda_{n})^{2}}+o(\frac{1}{(\log\lambda_{n})^{2}})$

It is easy to see that we get (2.3) from this if $L$ is not specified. In the

following we sketch how to get

$L=-8\pi\Lambda^{k}+2(3\log 2-1)$_.

The constant $L$ in (4.1) comes from two formula:

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where $d_{j}$ is the number given in (2.2), and

$\{\frac{1}{\mu_{n}^{k}}-u_{n}(x_{j,n})\}\gamma_{j,n}^{0}+16\pi c_{j}^{k}=-(8\pi)^{2}\sum_{i=1}^{m}g_{ji}c_{i}^{k}+o(1)$, (4.3)

where

$g_{ji}=\{\begin{array}{ll}\sum_{1<h\leq m}G(\kappa_{j}, \kappa_{h}) , for j=i,-h\neq j -G(\kappa_{j}, \kappa_{i}) , for j\neq i.\end{array}$

Concerning this point, previously we used

$u_{n}(x_{j,n})=-2\log\lambda_{n}+O(1)$

and

$\{\frac{1}{\mu_{n}^{k}}-u_{n}(x_{j,n})\}\gamma_{j,n}^{0}=O(1)$

.

Here we eliminate $u_{n}(x_{j,n})$ from these and get

$\{\frac{1}{\mu_{n}^{k}}+2\log\lambda_{n}+O(1)\}\gamma_{j,n}^{0}=O(1)$.

We know that $\gamma_{j,n}^{0}arrow 8\pi c_{j}^{k}$ and there exists at least one $j=1,$

$\cdots,$$m$ such

that $c_{j}^{k}\neq 0$ since $\mathfrak{c}^{k}=(c_{1}^{k}, \ldots, c_{m}^{k})\neq 0$

.

Therefore we get

$\frac{1}{\mu_{n}^{k}}=-2\log\lambda_{n}+O(1)$.

Similarly, eliminating $u_{n}(x_{j,n})$ from (4.2) and (4.3), we get

$\{\frac{1}{\mu_{n}^{k}}+2\log\lambda_{n}+2\log d_{j}+o(1)\}\gamma_{j,n}^{0}+16\pi c_{j}^{k}=-(8\pi)^{2}\sum_{i=1}^{m}g_{ji}c_{i}^{k}+o(1)$,

that is,

$\frac{1}{\mu_{n}^{k}}=-2\log\lambda_{n}-2-2\log d_{j}-\frac{8\pi\sum_{i--1}^{m}g_{ji^{C_{i}^{k}}}}{c_{j}^{k}}+o(1)$

$=-2 \log\lambda_{n}+2(3\log 2-1)-\frac{8\pi\sum_{i--1}^{m}h_{ji}c_{i}^{k}}{c_{j}^{k}}+o(1)$.

Here we

assume

$c_{j}^{k}\neq 0$ for simplicity. Since this formula exists for each

$j=1,$ $\cdots,$ $m$. We are able to get

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for $j\neq l$

.

This

means

that there exists

a

constant $\Lambda^{k}$

such that

$\sum_{i=1}^{m}h_{ji}c_{i}^{k}=\Lambda^{k}c_{j}^{k}$

for every $j=1,$ $\cdots,$$m$, that is,

$\Lambda^{k}$

is an eigenvalue of the matrix $(h_{ji})$,

see

[7] for details. Obviously $\mathfrak{c}^{k}=(c_{1}^{k}, \ldots, c_{m}^{k})$ is

an

eigenvector of $(h_{ji})$.

Finally

we are

able to conclude that $\Lambda^{k}$

is the k-th eigenvalue of $(h_{ji})$

because $\mu_{n}^{1}\leq\cdots\leq\mu_{n}^{k}.$

4.1 Derivation of

(4.2)

The formula (4.2) was essentially proved by C. C. Chen and C.-S. Lin [2,

Estimate $D$] from the Green’s representation formula:

$u_{n}(x_{j,n})= \int_{\Omega}G(x_{j,n}, y)\lambda_{n}e^{u_{n}(y)}dy$

$= \frac{1}{2\pi}\int_{B_{R}(x_{j,n})}\log|x_{j,n}-y|^{-1}\lambda_{n}e^{u_{n}(y)}dy$

$+ \int_{B_{R}(x_{j,n})}K(x_{j,n}, y)\lambda_{n}e^{u_{n}(y)}dy$

$+ \sum_{1\leq i\leq m,i\neq j}\int_{B_{R}(x_{i,n})}G(x_{j,n}, y)\lambda_{n}e^{u_{n}(y)}dy$

$+ \int_{\Omega\backslash \bigcup_{i=1}^{m}B_{R}(x_{i,n})}G(x_{j,n}, y)\lambda_{n}e^{u_{n}(y)}dy$

$=- \frac{\sigma_{j,n}}{2\pi}\log\delta_{j,n}+\frac{1}{2\pi}\int_{B_{\frac{R}{\delta_{j,n}}}(0)}\log|\tilde{y}|^{-1}e^{\tilde{u}_{j_{)}n}(\overline{y})}d\tilde{y}$

$+8 \pi\{R(x_{j,n})+\sum_{1\leq i\leq m ,i\neq j}G(x_{j,n}, x_{i,n})\}+o(1)$,

where

$\sigma_{j,n}:=\lambda_{n}\int_{B_{R}(x_{j,n})}e^{u_{n}}dx.$

We are able to confirm that

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and

$8 \pi\{R(x_{j,n})+\sum_{1\leq i\leq m ,i\neq j}G(x_{j,n}, x_{i,n})\}arrow 2\log(8d_{j})$

.

(4.5)

On the other hand, we know $\sigma_{j,n}arrow 8\pi$ from Theorem 1.1. Therefore

we get the following formula from the relation (3.1):

$u_{n}(x_{j,n})=- \frac{\sigma_{j,n}}{\sigma_{j,n}-4\pi}\log\lambda_{n}-2\log d_{j}+o(1)$ (4.6)

$=-2 \log\lambda_{n}+\frac{\sigma_{j,n}-8\pi}{\sigma_{j,n}-4\pi}\log\lambda_{n}-2\log d_{j}+o(1)$. (4.7)

Toget (4.2), we need to knowmore precise behavior of$\sigma_{j,n}$ along$\lambda_{n}arrow 0.$

To this purpose the following one obtained in [2, (3.56)] is suﬃcient:

$\sigma_{j,n}=8\pi+o(\lambda_{n})$ . (4.8)

We note that a weaker version

$\sigma_{j,n}=8\pi+o(\lambda^{\frac{1}{n2}})$ ,

which is also suﬃcient for our purpose, can be obtained rather easily, see

[13].

4.2 Derivation

of (4.3)

To get the formula (4.3) we use the Green’s theorem for $u_{n}$ and $\frac{v_{n}^{k}}{\mu_{n}^{k}}$ around

$B_{R}(x_{j,n})$:

$\int_{\partial B_{R}(x_{j,n})}\{\frac{\partial u_{n}}{\partial\nu}\frac{v_{n}^{k}}{\mu_{n}^{k}}-u_{n}\frac{\partial}{\partial\nu}(\frac{v_{n}^{k}}{\mu_{n}^{k}})\}d\sigma=\int_{B_{R}(x_{j,n})}\{\triangle u_{n}\frac{v_{n}^{k}}{\mu_{n}^{k}}-u_{n}\triangle\frac{v_{n}^{k}}{\mu_{n}^{k}}\}dx$

(4.9) The choice of $u_{n}$ and

$\frac{v_{n}^{k}}{k}$

seems

to be a kind of trick. Indeed we know the

$\mu_{n}$

behaviors of $u_{n}$ and $\frac{v}{\mu}2_{k^{Z},n}k$ far from $S=\{\kappa_{1}, \cdots, \kappa_{m}\}$

, see (1.4) and (2.7).

Therefore the left-hand side of (4.9) has limit in the process $narrow\infty.$

In fact, we have

$\int_{\partial B_{R}(x_{j,n})}\{\frac{\partial u_{n}}{\partial\nu}\frac{v_{n}^{k}}{\mu_{n}^{k}}-u_{n}\frac{\partial}{\partial v}(\frac{v_{n}^{k}}{\mu_{n}^{k}})\}d\sigma$

$m$ $m$

$arrow(8\pi)^{2}\sum\sum c_{i}^{k}\int_{\partial B_{R}(\kappa_{j})}\{\frac{\partial}{\partial\nu}G(x, \kappa_{h})G(x, \kappa_{i})-G(x, \kappa_{h})\frac{\partial}{\partial\nu}G(x, \kappa_{i})\}d\sigma.$

$h=1i=1$

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It is easy to

see

that

$\int_{\partial B_{R}(\kappa_{j})}\{\frac{\partial}{\partial\nu}G(x, \kappa_{h})G(x, \kappa_{i})-G(x, \kappa_{h})\frac{\partial}{\partial\nu}G(x, \kappa_{i})\}d\sigma$

$=\{\begin{array}{ll}0, h=i-G(\kappa_{j}, \kappa_{i})\delta_{h}^{j}+G(\kappa_{j}, \kappa_{h})\delta_{j}^{i}, h\neq i,\end{array}$

where $\delta_{a}^{b}=1$ if $a=b$ and $\delta_{a}^{b}=0$ else.

Therefore, from (4.10)

we

have

$\int_{\partial B_{R}(x_{j_{)}n})}\{\frac{\partial u_{n}}{\partial\nu}\frac{v_{n}^{k}}{\mu_{n}^{k}}-u_{n}\frac{\partial}{\partial\nu}(\frac{v_{n}^{k}}{\mu_{n}^{k}})\}d\sigma$

$=(8 \pi)^{2}\sum_{h=1}^{m}\sum_{1\leq i\leq m,i\neq h}c_{i}^{k}\{-G(\kappa_{j},\kappa_{i})\delta_{h}^{j}+G(\kappa_{j}, \kappa_{h})\delta_{j}^{i}\}+o(1)$

$=(8 \pi)^{2}\sum_{i=1}^{m}g_{ji}c_{i}^{k}+o(1)$.

On the other hand, we are able to apply the scaling argument to the

right-hand side of (4.9). Indeed the following holds:

$\int_{B_{R}(x_{j,n})}\{\Delta u_{n}\frac{v_{n}^{k}}{\mu_{n}^{k}}-u_{n}\triangle\frac{v_{n}^{k}}{\mu_{n}^{k}}\}dx$ $=- \frac{1}{\mu_{n}^{k}}\lambda_{n}\int_{B_{R}(x_{j,n})}e^{u_{n}}v_{n}^{k}dx+\lambda_{n}\int_{B_{R}(x_{j,n})}e^{u_{n}}v_{n}^{k}u_{n}dx$ $=- \frac{1}{\mu_{n}^{k}}\lambda_{n}\int_{B_{R}(x_{j,n})}e^{u_{n}}v_{n}^{k}dx+u_{n}(x_{j,n})\lambda_{n}\int_{B_{R}(x_{j_{)}n})}e^{u_{n}}v_{n}^{k}dx$ $+ \int_{B_{\frac{R}{\delta_{j,n}}}(0)}e^{\overline{u}_{j,n}}\tilde{v}_{j,n}^{k}\tilde{u}_{j,n}d\tilde{x}$ $=-( \frac{1}{\mu_{n}^{k}}-u_{n}(x_{j,n}))\gamma_{j,n}^{0}+\int_{B_{\frac{R}{\delta_{j_{)}n}}}(0)}e^{\tilde{u}_{j,n}}\tilde{v}_{j,n}^{k}\tilde{u}_{j,n}d\tilde{x}.$ Here $\int_{B_{\frac{R}{0_{j,n}}}(0)}e^{\overline{u}_{j_{)}n}}\tilde{v}_{j,n}^{k}\tilde{u}_{j,n}d\tilde{x}arrow\int_{\mathbb{R}^{2}}e^{U}V_{j}^{k}Ud\tilde{x}=-16\pi c_{j}^{k}.$ Consequently we get (4.3).

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5Examples of

$\mathfrak{c}^{k}=$

$(c_{1}^{k}, , c_{m}^{k})\neq 0$

Let us fix an integer $m\geq 2$ and $\Omega$

be an annulus $\{x\in \mathbb{R}^{2};(0<)a<|x|<1\}.$

In [12] there was constructed a $m$-mode solution $u_{n}$ to (1.1), i.e. a solution

which is invariant with respect to a rotation of $\frac{2\pi}{m}$ in $\mathbb{R}^{2},$

$u(r, \theta)=u(r, (\theta+\frac{2\pi}{m}))$

This solution blows-up at $m$ points $\kappa_{1}=\cdots=\kappa_{m}$ which are located on a

circle concentric with the annulus and

are

verticesof

a

regular polygon with$m$

sides. So we can assume that $\kappa_{1}=(r_{0},0)$, $\kappa_{2}=r_{0}(\cos\frac{2\pi}{m}, \sin\frac{2\pi}{m})$ , . .

.

,_{$\kappa_{m}=$}

$r_{0}( \cos\frac{2(m-1)\pi}{m}, \sin\frac{2(m-1)\pi}{m})$ for some _{$r_{0}\in(a, 1)$}.

Since $G(x, \kappa_{1})$ is symmetricwith respect tothe

$x_{1}$-axis, weget $G(\kappa_{j}, \kappa_{1})=$

$G(\kappa_{m-j+2}, \kappa_{1})$, $j=2,$ _$m$. Similarly the value _{$G(\kappa_{i}, \kappa_{j})$} depends only

on the

distance between $\kappa_{i}$ and

$\kappa_{j}$. Since

$\Omega$ is an annulus,

the Robin function $R(x)$

is radial, so that $R(\kappa_{1})=\cdots=R(\kappa_{m})=R.$

Here weset $G_{i}$ $:=G(\kappa_{i}, \kappa_{1})$ and $R_{l}$ $:=R+4 \sum_{h=2}^{l}G_{h}$ for simplicity. Then

the matrix $h_{ij}$ becomes as follows:

for $m=2l(l=1,2, \cdots)$,

$(h_{ij})=(\begin{array}{lllllll}R_{l}+2G_{l+1} -G_{2} -G_{3} \cdots -G_{l+1} \cdots -G_{2}-G_{2} R_{l}+2G_{l+l} -G_{2} \cdots \cdots \cdots -G_{3}-G_{2}\cdots \cdots \cdots \cdots -G_{3} \cdots\cdots \cdots R_{l}+2G_{l+1}\end{array}),$

and for $m=2l+1(l=1,2, \cdots)$,

$(h_{ij})=(\begin{array}{llllllll}R_{l} -G_{2} -G_{3} \cdots -G_{l} -G_{l} \cdots -G_{2}-G_{2} R_{l} -G_{2} \cdots \cdots \cdots \cdots -G_{3}-G_{2}\cdots -G_{3} \cdots\cdots \cdots\cdots \cdots\cdots \cdots\cdots \cdots-G_{2} \cdots R_{l}\end{array}),$

A straightforward computation _{shows the following facts:}

$\bullet m=3$

$\lambda_{1}=R+2G_{2},$ $\mathfrak{c}_{1}=(1,1,1)$.

$\lambda_{2}=\lambda_{3}=R+5G_{2},$ $\mathfrak{c}_{2}=(1, -1,0)$, $\mathfrak{c}_{3}=(1,0, -1)$. $\bullet m=4$

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$\lambda_{2}=\lambda_{3}=R+4G_{2}+3G_{3},$ $\mathfrak{c}_{2}=(1,0, -1,0)$, $\mathfrak{c}_{3}=(0,1,0, -1)$.

$\lambda_{4}=R+6G_{2}+G_{3},$ $\mathfrak{c}_{4}=(1, -1,1, -1)$.

$\bullet m=5$

$\lambda_{1}=R+2G_{2}+2G_{3},$ $c_{1}=(1,1,1,1,1)$,

$\lambda_{2}=\lambda_{3}=R+\frac{9-\sqrt{5}}{2}G_{2}+\frac{9+\sqrt{5}}{2}G_{3},$

$\mathfrak{c}_{2}=(1, \frac{-1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2}, -1,0)$,

C3 $=(1, -1, \frac{-1-\sqrt{5}}{2},0, \frac{1+\sqrt{5}}{2})$,

$\lambda_{4}=\lambda_{5}=R+\frac{9+\sqrt{5}}{2}G_{2}+\frac{9-\sqrt{5}}{2}G_{3}.$

$\mathfrak{c}_{4}=(1, \frac{-1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}, -1,0)$,

C5 $=(1, -1, \frac{-1+\sqrt{5}}{2},0, \frac{1-\sqrt{5}}{2})$. $\bullet m=6$ $\lambda_{1}=R+2G_{2}+2G_{3}+G_{4},$ $\mathfrak{c}_{1}=(1,1,1,1,1,1)$, $\lambda_{2}=\lambda_{3}=R+3G_{2}+5G_{3}+3G_{4},$ $\mathfrak{c}_{2}=(1,0, -1, -1,0,1)$, $\mathfrak{c}_{3}=(1,1,0, -1, -1,0)$ $\lambda_{4}=\lambda_{5}=R+5G_{2}+5G_{3}+G_{4},$ $\mathfrak{c}_{4}=(1, -1,0, -1,1,0)$, $\mathfrak{c}_{5}=(1,0,1, -1,0,1)$ $\lambda_{6}=R+6G_{2}+2G_{3}+3G_{4},$ $\mathfrak{c}_{6}=(1, -1,1, -1,1, -1)$ $\bullet$

. . .

In general, it is easy to see that the first eigenvalue of $(h_{ij})$ is $\Lambda^{1}=R+$

$2 \sum_{h=2}^{l}G_{h}+G_{l+1}$ for $m=2l$ and $R+2 \sum_{h=2}^{l}G_{h}$ for $m=2l+1$ which is

simple. It is easy to see that the eigenspace corresponding to $\Lambda^{1}$

is spanned by $\mathfrak{c}^{1}=(1,1, \cdots, 1)$.

Unfortunatelywe are not yet able to get further informationon the

multi-plicity oftheeigenvalues even for these

cases

and we will leave this for future

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